Question answer I don't understand

  • Thread starter Thread starter Genericcoder
  • Start date Start date
AI Thread Summary
The discussion revolves around calculating the probability that a defective fuse came from a specific production line after a customer tested a lot. The production lines typically produce 2% defective fuses, but line 1 had a mechanical issue, resulting in 5% defectives. The calculations provided show how to determine the probabilities using Bayes' theorem, with specific values for P(A|B) and P(B-). A participant clarifies a common confusion regarding the use of the formulas, suggesting that the events A and B may be flipped from standard textbook presentations. Ultimately, the example illustrates that approximately 38% of the defective fuses could be traced back to line 1.
Genericcoder
Messages
130
Reaction score
0
An electronic fuse is produced by five production lines in a manufacturing operation.
The fuses are costly, are quite reliable, and are shipped to suppliers in 100-unit lots.
Because testing is destructive, most buyers of the fuses test only a small number of
fuses before deciding to accept or reject lots of incoming fuses.
All five production lines produce fuses at the same rate and normally produce
only 2% defective fuses, which are dispersed randomly in the output. Unfortunately,
production line 1 suffered mechanical difficulty and produced 5% defectives during
the month of March. This situation became known to the manufacturer after the fuses
had been shipped.Acustomer received a lot produced in March and tested three fuses.
One failed. What is the probability that the lot was produced on line 1? What is the
probability that the lot came from one of the four other lines?

Let B denote the event that a fuse was drawn from line 1 and let A denote the event
that a fuse was defective. Then it follows directly that

P(B) = 0.2 and P(A|B) = 3(.05)(.95)^2 = .135375.

Similarly,
P(B-) = 0.8 and P(A|B-) = 3(.02)(.98)2 = .057624.



P(A) = P(A|B)P(B) + P(A|B-)P(B-)
= (.135375)(.2) + (.057624)(.8) = .0731742.


P(B|A) = P(B & A) / P(A) = P(A|B)*P(B) / P(A) = (.135375)(.2) / .0731742 = 0.37

Wat I don't understand here how did he get those values for
P(A|B) and P(B-)..? shouldn't P(B | A) = P(A & B) / P(B) = 0.05/0.2

P(B) = 1/5 = 0.2. same logic for P(A|B-) I don't understand this if someone could explain this more clearly.
 
Physics news on Phys.org
Genericcoder said:
shouldn't P(B | A) = P(A & B) / P(B) = 0.05/0.2

It should be P(A) in the denominator, not P(B) like you have written. You may be confused because the formula will typically be presented as giving P(A | B), so your A and B are flipped from what the standard formula in a textbook would read.
 
oh i see
 
Another way of looking at it: Imagine that every line produces 1000 items, for a total of 5000 items. Lines 2 through 4 have 2% bad: a total of .02(4000)= 80 bad items. Line one produces 5% bad, a total of 50 bad items. That is, out of a total of 80+ 50= 130 bad items, 50, or 50/130 or about 38% came from line 1.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
Back
Top